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In a composite rod, when two rods of different lengths and of the same area are joined end to end. If $K$ is the coefficient of thermal conductivity of composite rod, then $\left( {\frac{{{\ell _1} + {\ell _2}}}{K}} \right)$ is equal to
$\frac{{{\ell _1}}}{{{K_1}}} - \frac{{{\ell _2}}}{{{K_2}}}$
$\frac{{{\ell _1}}}{{{K_2}}} - \frac{{{\ell _2}}}{{{K_1}}}$
$\frac{{{\ell _1}}}{{{K_1}}} + \frac{{{\ell _2}}}{{{K_2}}}$
$\frac{{{\ell _1}}}{{{K_2}}} + \frac{{{\ell _2}}}{{{K_1}}}$
Solution

$\mathrm{R}_{\mathrm{eq}^{-}}=\mathrm{R}_{1}+\mathrm{R}_{2}$
$\left(\frac{\ell_{1}+\ell_{2}}{\mathrm{K}_{\mathrm{eq}} \mathrm{A}}\right)=\frac{\ell_{1}}{\mathrm{K}_{1} \mathrm{A}}+\frac{\ell_{2}}{\mathrm{K}_{2} \mathrm{A}}$
$\frac{\ell_{1}+\ell_{2}}{\mathrm{K}}=\frac{\ell_{1}}{\mathrm{K}_{1}}+\frac{\ell_{2}}{\mathrm{K}_{2}}$